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Given $A,B\in {\Bbb Z}^2$, write $A \leftrightarrows B$ if the interior of the line segment AB misses ${\Bbb Z}^2$.

For $r>0$, define $S_r:=\{ \{A, B\} | A,B\in {\Bbb Z}^2,||A||<r,||B||<r, | ||A||-||B|| |<1 {\rm\ and\ } A \leftrightarrows B \}\ .$

A little calculus gives the equivalence of $\zeta(2)=\pi^2/6$ and $$\lim_{r\to\infty} \frac{|S_r|}{(2r)^3} = 1\ .$$

Of course $(2r)^3$ counts lattice points in a cube $C_{r}:=[-r,r)^3$.

Question: Does there exist an approximately bijective proof of this limit (or some variant), one that matches most of $S_r$ with most of $C_r$?

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2  
Do you have any reason to think there should be? –  Greg Martin May 2 '12 at 4:05
5  
@Greg Martin 1) There do exist reasonably elementary evaluations of $\zeta(2)$ (no Fourier analysis, no contour integration, etc.) so perhaps it's possible to unwind one of them. 2) Any combinatorialist presented with a combinatorial identity naturally wonders whether a bijective proof exists - I don't know of any serious candidates for natural combinatorial facts with no natural proofs. –  David Feldman May 2 '12 at 5:46
2  
Interesting question. Can you spell out the calculus a bit? I see various starts but not a clear path. –  Aaron Meyerowitz May 2 '12 at 7:11
1  
Interesting question, does this generalize to $\zeta(2k)$? –  user22202 May 2 '12 at 7:16
3  
Roughly (and not rigorously), each pair $A,B$ contains a first point reading clockwise, say $A$. The condition $|\,||A||−||B||\,|<1$ on $B$ in ${\Bbb R^2}$ locates $B$ in a half-annulus of area about $2\pi||A||$. This half-annulus contains a number of lattice points about equal to it's area, and of those, about $1/\zeta(2)$ satisfy $A⇆B$. Now do a double integral for the possible locations of $A$, first by norm, then by angle. The proportion of points where $A⇆B$ may not behave so will in a thin region, but some integration by parts makes everything legal. –  David Feldman May 2 '12 at 22:57

1 Answer 1

Here's a rough mapping using induction principles. We are aiming to prove that $|S_{r+2}|-|S_r|\approx 6r^2+12r+8$. We take the area of the annulus, which is $\pi(r+2)^2-\pi r^2=\pi(4r+4)$. We claim that the number of lattice points inside this region is therefore about $\pi(4r+4)$, and using triangular numbers, the number of actual lines is approximately $\pi^2(4r+4)(4r+5)/2$, and we remove a factor of $\zeta(2)$ to allow for the coprimality condition, leaving $3(4r+4)(4r+5)=3(16r^2+36r+20)$, which is not a bad approximation to the $6r^2+12r+8$ we require.

double ring

The double ring structure comes from the introduction of a new ring. We get both (new ring $\to$ new ring) and (new ring $\to$ previous ring) entries. The rarifaction of the previous ring leads to the final smoothing of the polynomial, so all we now need to do is scale the factors involved.

I wrote some reasonable JavaScript to test this:

<!DOCTYPE html>
<html>
<body>
<span id='s'></span>
</body>
<script>


function dist(x,y) {
return Math.sqrt((x[0]-y[0])*(x[0]-y[0])+(x[1]-y[1])*(x[1]-y[1]));
}

function gcd(a,b) {
var x,y;
x=Math.abs(a[0]-b[0]);
y=Math.abs(a[1]-b[1]);
if ((x==0 && y>1) || (x>1 &&  y==0)) return 0;
var g,mx=1;
for (g=2;g<=Math.max(x,y)/2;g++) if (x%g==0 && y%g==0) mx=g;
return mx;
}

for (d=0;d<=25;d++) {
c=0;
for (i=-d;i<=d;i++)
for (j=-d;j<=d;j++)
for (a=-d;a<=d;a++)
for (b=-d;b<=d;b++) {
if (a<i) continue;
if (a==i && b<=j) continue;
if (dist([i,j],[0,0])<=d && dist([a,b],[0,0])<=d && (i!=a || j!=b) && Math.abs(dist([i,j],[0,0])-dist([a,b],[0,0]))<1 && gcd([i,j],[a,b])==1) c++;
//if (d==2) console.log([i,j]+'..'+[a,b]);}
}
console.log(c+'..'+Math.pow(2*d,3));
}

</script>
</html>
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10  
Unless I'm mistaken, isn't the point of David Feldman's question that one wants to avoid using the fact $\zeta(2)=\pi^2/6$, which you seem to be using in your argument? –  Yemon Choi May 6 at 21:06
    
@YemonChoi; I can't see where he argues that. With coprimes, I can't see how you are going to avoid it. –  Jon Mark Perry May 7 at 5:28
7  
Look at the title of his question. I believe David Feldman wants to prove the limit formula he states, in order to deduce that $\zeta(2)=\pi^2/6$. So using the latter fact in proving the limit would, for the purposes of this question, be circular. –  Yemon Choi May 7 at 9:49

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